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Hunter-Best [27]
3 years ago
15

The velocity component with which a projectile covers certain horizontal distance is maximum at the moment of? a) Hitting the gr

ound b) Highest point c) Projection d) None of these
Physics
1 answer:
Crazy boy [7]3 years ago
3 0
A) hitting the ground
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When electrons are accelerated by 2450v in an electron microscope they will have wavelengths of
Sholpan [36]
I think the answer is A I’m
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5 0
2 years ago
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
ArbitrLikvidat [17]

Answer: Here is the complete question:

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 103 N/C. If the ball is in equilibrium when the string makes a 30 angle with the vertical, what is the net charge on the ball?

Answer: The charge on the ball is 5.71 × 10^-4 C

Explanation:

Please see the attachments below

5 0
3 years ago
The water in the plumbing in a house is at a gauge pressure of 300,000 pa. What force does this cause on the top of the tank ins
liraira [26]

Answer:

60 000 N

Explanation:

1 pa = 1 N/m^2

you have 300 000 of these   = 300 000 N /m^2

    but only an area of .2 m^2

                    300 000 N / m^2  * .2 m^2 = 60 000 N

3 0
1 year ago
A 10 N force and a 15 N force are acting from a single point in opposite directions. What additional force must be added to prod
AleksAgata [21]

Answer:

5 N acting in the same direction as the 10 N force

Explanation:

10+5=15

15=15

7 0
3 years ago
1. Determine the image distance in each of the following.
miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

1

=

f

1

v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

1

v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

1

+

∞

1

=

15

1

v=+15cm

The image is formed at a distance of 15cm in left of mirror

6 0
3 years ago
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